If the zeroes of the polynomial \(p(x) = 2x^2 - 5x + (k-2)\) are reciprocals of each other, then we can use the fact that the sum of the roots of a quadratic equation \(ax^2 + bx + c = 0\) is \(-b/a\) and the product of the roots is \(c/a\).
In this case, if \(a = 2x^2\), \(b = -5x\), and \(c = k-2\), we have:
Sum of roots: \(-b/a = -(-5x)/2x^2 = 5/2x\)
Product of roots: \(c/a = (k-2)/2x^2\)
Since the roots are reciprocals of each other, their product should be 1:
\((k-2)/2x^2 = 1\)
Now, solving for \(k\):
\(k - 2 = 2x^2\)
\(k = 2x^2 + 2\)
The value of \(k\) in terms of \(x\) is \(k = 2x^2 + 2\).
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If the zeroes of the polynomial \(p(x) = 2x^2 - 5x + (k-2)\) are reciprocals of each other, then we can use the fact that the sum of the roots of a quadratic equation \(ax^2 + bx + c = 0\) is \(-b/a\) and the product of the roots is \(c/a\).
In this case, if \(a = 2x^2\), \(b = -5x\), and \(c = k-2\), we have:
Sum of roots: \(-b/a = -(-5x)/2x^2 = 5/2x\)
Product of roots: \(c/a = (k-2)/2x^2\)
Since the roots are reciprocals of each other, their product should be 1:
\((k-2)/2x^2 = 1\)
Now, solving for \(k\):
\(k - 2 = 2x^2\)
\(k = 2x^2 + 2\)
The value of \(k\) in terms of \(x\) is \(k = 2x^2 + 2\).